Question

The function f is defined by f(x) = x^2 - kx + 3, where the domain of f is the set of all real numbers and k is a non-zero constant. Given that the range of f is [-1 , infinity) and that f^-1 exists if the domain of f is restricted to the set of negative real numbers, find the value of k.

Answer 1

\[-1\le x^2-kx+3< \infty \]

Answer 2

we don't have any idea about infinity
so we will go with
\[-1 \le x^2-kx+3\]

Answer 3

x^2-kx+3+1>=0
x^2-kx+4>=0

Answer 4

it is quadratic equation

Answer 5

so for it to be >=0
D>=0
k^2-4*1*4>=0
k^2-16>=0
(k-4)(k+4)>=0

Answer 6

k>=4 or k<=-4

Answer 7

The answer given is only k = 4. Any idea why?

Answer 8

Nevermind, I managed to solve it.

Since f(x) is a positive quadratic curve, it has a minimum point.
f(x) = x^2 - kx + 3
f'(x) = 2x - k
For f(x) to be minimum, f'(x) = 0.
2x - k = 0
x = k / 2
Since minimum f = -1,
f(k / 2) = -1
(k / 2)^2 - k(k / 2) + 3 = -1
k^2 / 4 - k^2 / 2 + 4 = 0
k^2 - 2k^2 + 16 = 0
k^2 - 16 = 0
k = 4 or k = -4

Next, the domain of f is restricted to (-inf , 0).
Assume k = -4,
f'(x) = 2x + 4, x < 0
Since f'(x) changes from positive to negative, f(x) is not strictly increasing or decreasing as x increases. Thus, f^-1 does not exists, which contradicts the question.

Assume k = 4,
f'(x) = 2x - 4, x < 0
Since f'(x) remains negative, f(x) is strictly decreasing as x increases. Thus, f^-1 exists.

So, k = 4.